{"segment":[{"part":[{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["-10"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/2.jpg' \/><\/center> \u0110\u1ec3 ph\u00e2n th\u1ee9c $\\dfrac{5x-2}{x^2+10x}$ c\u00f3 ngh\u0129a th\u00ec $\\left\\{ \\begin{align} & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ & x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} \\\\ \\end{align} \\right.$ ","explain":" \u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u00e2n th\u1ee9c c\u00f3 ngh\u0129a l\u00e0:<br\/>$x^2+10x\\ne 0$ <br\/> $\\Rightarrow x(x+10) \\ne 0$ <br\/> $\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x+10 \\ne 0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne -10 \\\\ \\end{aligned} \\right.$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$ v\u00e0 $-10.$ <\/span><\/span> "}],"id_ques":251},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'> K\u1ebft qu\u1ea3 r\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $ Q=\\left( \\dfrac{2ab}{{{a}^{2}}-{{b}^{2}}}+\\dfrac{a-b}{2a+2b} \\right)\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $ l\u00e0 _input_ <\/span> ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $a \\ne \\pm b \\ne 0$ <br\/> Ta c\u00f3:<br\/> $ Q=\\left( \\dfrac{2ab}{{{a}^{2}}-{{b}^{2}}}+\\dfrac{a-b}{2a+2b} \\right)\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $<br\/>$ =\\left[ \\dfrac{2ab}{\\left( a+b \\right)\\left( a-b \\right)}+\\dfrac{a-b}{2\\left( a+b \\right)} \\right]\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $<br\/>$ =\\left[ \\dfrac{2ab.2}{2\\left( a+b \\right)\\left( a-b \\right)}+\\dfrac{{{\\left( a-b \\right)}^{2}}}{2\\left( a+b \\right)\\left( a-b \\right)} \\right]\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $<br\/>$ =\\dfrac{4ab+{{\\left( a-b \\right)}^{2}}}{2\\left( a+b \\right)\\left( a-b \\right)}\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $<br\/>$ =\\dfrac{4ab+{{a}^{2}}-2ab+{{b}^{2}}}{2\\left( a+b \\right)\\left( a-b \\right)}\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $<br\/>$ =\\dfrac{{{\\left( a+b \\right)}^{2}}}{2\\left( a+b \\right)\\left( a-b \\right)}\\cdot \\dfrac{2a}{a+b}$$+\\dfrac{b}{b-a} $<br\/>$ =\\dfrac{\\left( a+b \\right)2a}{2\\left( a-b \\right)\\left( a+b \\right)}+\\dfrac{b}{b-a} $<br\/>$ =\\dfrac{a}{a-b}+\\dfrac{b}{b-a} $<br\/>$ =\\dfrac{a-b}{a-b} $<br\/>$ =1 $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1.$ <\/span><\/span> "}],"id_ques":252},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["2"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'> T\u00ecm $x\\in \\mathbb{Z}$ \u0111\u1ec3 $C = \\dfrac{x}{x-1}\\in \\mathbb{Z}$ <br\/> \u0110\u00e1p \u00e1n: $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$ <\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x \\ne 1$ <br\/> Ta c\u00f3:<br\/> $C=\\dfrac{x}{x-1}=\\dfrac{x-1+1}{x-1}$$=1+\\dfrac{1}{x-1}$ <br\/> \u0110\u1ec3 $C\\in \\mathbb{Z}$ th\u00ec $\\dfrac{1}{x-1}\\in \\mathbb{Z}$ <br\/> Khi \u0111\u00f3 $x-1\\in \u01af\\left( 1 \\right)$ , m\u00e0 $\u01af\\left( 1 \\right)=\\left\\{ -1;1 \\right\\}$ <br\/> Ta c\u00f3 b\u1ea3ng sau:<br\/> <table> <tr> <th>$x-1$<\/th> <th>$-1$<\/th> <th>$1$<\/th> <\/tr> <tr> <td>$x$<\/td> <td>$0$<\/td> <td>$2$<\/td> <\/tr><\/table> <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0; 2.$ <\/span><\/span> "}],"id_ques":253},{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/3.jpg' \/><\/center> M\u1eabu th\u1ee9c chung c\u1ee7a c\u00e1c ph\u00e2n th\u1ee9c $\\dfrac{x+a}{a{{b}^{3}}x};\\,\\,\\dfrac{x+b}{{{a}^{2}}{{b}^{2}}x};\\,\\,\\dfrac{a+b}{{{x}^{2}}{{b}^{3}}}$ l\u00e0","select":["A. $ab^3x$ ","B. $a^2b^3x^2$ ","C. $a^3b^3x^2$","D. $abx$"],"hint":" T\u00ecm t\u00edch c\u1ee7a c\u00e1c nh\u00e2n t\u1eed c\u00f3 trong m\u1eabu c\u00e1c ph\u00e2n th\u1ee9c v\u1edbi s\u1ed1 m\u0169 cao nh\u1ea5t. ","explain":"<span class='basic_left'> $ BCNN(1;1) = 1$<br\/> Th\u1eeba s\u1ed1 ri\u00eang $a$ c\u00f3 s\u1ed1 m\u0169 cao nh\u1ea5t l\u00e0 $2$<br\/> C\u00e1c th\u1eeba s\u1ed1 ri\u00eang $b$ v\u00e0 $x$ c\u00f3: <br\/> S\u1ed1 m\u0169 cao nh\u1ea5t c\u1ee7a $b$ l\u00e0 $3$<br\/> S\u1ed1 m\u0169 cao nh\u1ea5t c\u1ee7a $x$ l\u00e0 $2.$<br\/> Suy ra m\u1eabu th\u1ee9c chung l\u00e0: $a^2b^3x^2$ <span><br\/> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 B.<\/span> <br\/> <br\/><b> L\u01b0u \u00fd:<\/b> <br\/> <i> Mu\u1ed1n t\u00ecm m\u1eabu th\u1ee9c chung c\u1ee7a c\u00e1c ph\u00e2n th\u1ee9c: <br\/> 1. Ph\u00e2n t\u00edch c\u00e1c m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed <br\/> 2. M\u1eabu th\u1ee9c chung c\u1ea7n t\u00ecm l\u00e0 m\u1ed9t t\u00edch m\u00e0 c\u00e1c nh\u00e2n t\u1eed \u0111\u01b0\u1ee3c ch\u1ecdn nh\u01b0 sau: <br\/> + H\u1ec7 s\u1ed1 c\u1ee7a m\u1eabu th\u1ee9c chung b\u1eb1ng b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a h\u1ec7 s\u1ed1 c\u00e1c m\u1eabu th\u1ee9c. <br\/> + V\u1edbi m\u1ed7i l\u0169y th\u1eeba c\u1ee7a c\u00f9ng m\u1ed9t bi\u1ec3u th\u1ee9c c\u00f3 m\u1eb7t trong c\u00e1c m\u1eabu th\u1ee9c, ta ch\u1ecdn l\u0169y th\u1eeba v\u1edbi s\u1ed1 m\u0169 l\u1edbn nh\u1ea5t. <\/i>","column":2}],"id_ques":254},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/3.jpg' \/><\/center> Gi\u00e1 tr\u1ecb c\u1ee7a ph\u00e2n th\u1ee9c $\\dfrac{x^2-1}{x^2+2x+1}$ b\u1eb1ng $0$ khi $x =$ _input_ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/><b> B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u00e2n th\u1ee9c.<br\/> <b> B\u01b0\u1edbc 2:<\/b> Cho t\u1eed th\u1ee9c b\u1eb1ng $0$ d\u1ec3 t\u00ecm $x$ <br\/> <b> B\u01b0\u1edbc 3:<\/b> K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 k\u1ebft lu\u1eadn. <br\/> <span class='basic_green'>Gi\u1ea3i<\/span><br\/> \u0110i\u1ec1u ki\u1ec7n c\u1ee7a ph\u00e2n th\u1ee9c l\u00e0: $x\\ne -1$ <br\/> Khi \u0111\u00f3:<br\/> $\\begin{aligned} & \\dfrac{{{x}^{2}}-1}{{{x}^{2}}+2x+1}=0 \\\\ & \\Rightarrow {{x}^{2}}-1=0 \\\\ &\\Rightarrow \\left( x+1 \\right)\\left( x-1 \\right)=0 \\\\ & \\Rightarrow\\left[ \\begin{aligned} & x+1=0 \\\\ & x-1=0 \\\\ \\end{aligned} \\right.\\Rightarrow \\left[ \\begin{aligned} & x=-1\\,\\,\\left( \\text{lo\u1ea1i} \\right) \\\\ & x=1\\,\\,\\,\\left( \\text{th\u1ecfa m\u00e3n} \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1.$ <\/span><\/span><\/span> "}],"id_ques":255},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank_random","correct":[[["5x"],["8"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ B=\\left( \\dfrac{2x+1}{x-1}+\\dfrac{8}{{{x}^{2}}-1}-\\dfrac{x-1}{x+1} \\right)$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $ <br\/> <br\/> <br\/> <b> C\u00e2u a: <\/b> R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $B$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: $B = \\dfrac{x^2+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}+\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{5}$<\/span> ","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{-1;1\\}$ <br\/> Ta c\u00f3: <br\/> $ B=\\left( \\dfrac{2x+1}{x-1}+\\dfrac{8}{{{x}^{2}}-1}-\\dfrac{x-1}{x+1} \\right)$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $<br\/>$ =\\left[ \\dfrac{\\left( 2x+1 \\right)\\left( x+1 \\right)}{{{x}^{2}}-1}+\\dfrac{8}{{{x}^{2}}-1}-\\dfrac{{{\\left( x-1 \\right)}^{2}}}{{{x}^{2}}-1} \\right]$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $<br\/>$ =\\dfrac{\\left( 2x+1 \\right)\\left( x+1 \\right)+8-{{\\left( x-1 \\right)}^{2}}}{{{x}^{2}}-1}$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $<br\/>$ =\\dfrac{2{{x}^{2}}+3x+1+8-{{x}^{2}}+2x-1}{{{x}^{2}}-1}$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $<br\/>$ =\\dfrac{{{x}^{2}}+5x+8}{{{x}^{2}}-1}$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $<br\/>$ =\\dfrac{{{x}^{2}}+5x+8}{5} $ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng \u0111\u1ec3 c\u00f3 k\u1ebft qu\u1ea3 r\u00fat g\u1ecdn l\u00e0 $\\dfrac{{{x}^{2}}+5x+8}{5}$. <\/span><\/span> "}],"id_ques":256},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t.<br\/> Kh\u1eb3ng \u0111\u1ecbnh d\u01b0\u1edbi \u0111\u00e2y \u0111\u00fang hay sai","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/10.jpg' \/><\/center> <span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $ B=\\left( \\dfrac{2x+1}{x-1}+\\dfrac{8}{{{x}^{2}}-1}-\\dfrac{x-1}{x+1} \\right)$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $ <br\/> <br\/> <br\/> <b> C\u00e2u b: <\/b> Ta kh\u1eb3ng \u0111\u1ecbnh \u0111\u01b0\u1ee3c $B > 0$ v\u1edbi m\u1ecdi $x\\ne \\{-1;1\\}$ <\/span> ","select":["\u0110\u00fang","Sai"],"explain":"<span class='basic_left'><span class='basic_left'> Theo c\u00e2u a tr\u00ean, ta c\u00f3: $ B =\\dfrac{{{x}^{2}}+5x+8}{5}$ <br\/> T\u1eed th\u1ee9c $= {{x}^{2}}+5x+8 \\\\ ={{x}^{2}}+2.x.\\dfrac{5}{2}+\\frac{25}{4}-\\dfrac{25}{4}+8 \\\\ ={{\\left( x+\\dfrac{5}{2} \\right)}^{2}}+\\dfrac{7}{4}\\,\\,>0\\,\\,\\forall x $ <br\/> M\u1eabu th\u1ee9c $= 5 > 0$ <br\/> Do \u0111\u00f3 $B > 0$ v\u1edbi m\u1ecdi $x\\ne \\{-1;1\\}$ <br\/> <span> <span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n l\u00e0 \u0110\u00fang.<\/span>","column":2}],"id_ques":257},{"title":"\u0110i\u1ec1n \u0111\u00e1p \u00e1n th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho ba c\u00e2u","temp":"fill_the_blank","correct":[[["7"],["20"]]],"list":[{"point":10,"width":50,"type_input":"","input_hint":["frac"],"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/10.jpg' \/><\/center>Cho bi\u1ec3u th\u1ee9c $ B=\\left( \\dfrac{2x+1}{x-1}+\\dfrac{8}{{{x}^{2}}-1}-\\dfrac{x-1}{x+1} \\right)$$\\cdot \\dfrac{{{x}^{2}}-1}{5} $ <br\/> <br\/> <br\/> <b> C\u00e2u c: <\/b> GTNN c\u1ee7a $B$ l\u00e0 <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":" T\u00ednh GTNN c\u1ee7a t\u1eed th\u1ee9c r\u1ed3i suy ra GTNN c\u1ee7a $B$","explain":"<span class='basic_left'> Theo c\u00e2u a tr\u00ean, ta c\u00f3: $B=\\dfrac{{{x}^{2}}+5x+8}{5} $<br\/> Theo c\u00e2u b tr\u00ean, ta ph\u00e2n t\u00edch \u0111\u01b0\u1ee3c t\u1eed th\u1ee9c th\u00e0nh: ${{x}^{2}}+5x+8$$={{\\left( x+\\dfrac{5}{2} \\right)}^{2}}+\\dfrac{7}{4}\\ge \\dfrac{7}{4}$ <br\/> Do \u0111\u00f3 GTNN c\u1ee7a t\u1eed th\u1ee9c l\u00e0 $\\dfrac{7}{4}$ <br\/> V\u1eady GTNN c\u1ee7a $B$ l\u00e0: $\\dfrac{\\dfrac{7}{4}}{5}=\\dfrac{7}{20}$ <br\/> D\u1ea5u '=' x\u1ea3y ra khi $\\left( x+\\dfrac{5}{2} \\right)^2=0$ hay $x=-\\dfrac{5}{2}$<\/span> "}],"id_ques":258},{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-1"],["-1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/16.jpg' \/><\/center> Gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $B=\\dfrac{{{\\left( x+2 \\right)}^{2}}}{x}\\cdot \\left( 1-\\dfrac{{{x}^{2}}}{x+2} \\right)$$-\\dfrac{{{x}^{2}}+6x+4}{x}$ l\u00e0: _input_ <br\/> D\u1ea5u \u201c=\u201d x\u1ea3y ra khi $x = $ _input_","explain":" <span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{0;-2\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & B=\\dfrac{{{\\left( x+2 \\right)}^{2}}}{x}\\cdot \\left( 1-\\dfrac{{{x}^{2}}}{x+2} \\right)-\\dfrac{{{x}^{2}}+6x+4}{x} \\\\ & =\\dfrac{{{\\left( x+2 \\right)}^{2}}}{x}\\cdot \\dfrac{\\left( x+2-{{x}^{2}} \\right)}{x+2}-\\dfrac{{{x}^{2}}+6x+4}{x} \\\\ & =\\dfrac{{{\\left( x+2 \\right)}^{2}}\\left( -{{x}^{2}}+x+2 \\right)}{x\\left( x+2 \\right)}-\\dfrac{{{x}^{2}}+6x+4}{x} \\\\ & =\\dfrac{\\left( x+2 \\right)\\left( -{{x}^{2}}+x+2 \\right)}{x}-\\dfrac{{{x}^{2}}+6x+4}{x} \\\\ & =\\dfrac{-{{x}^{3}}+{{x}^{2}}+2x-2{{x}^{2}}+2x+4-{{x}^{2}}-6x-4}{x} \\\\ & =\\dfrac{-{{x}^{3}}-2{{x}^{2}}-2x}{x} \\\\ & =-{{x}^{2}}-2x-2 \\\\ & =-\\left( {{x}^{2}}+2x+2 \\right) \\\\ & =-\\left( {{x}^{2}}+2x+1+1 \\right) \\\\ & =-\\left[ {{\\left( x+1 \\right)}^{2}}+1 \\right] \\\\ & =-1-{{\\left( x+1 \\right)}^{2}} \\\\ & Do\\,\\,{{\\left( x+1 \\right)}^{2}}\\ge 0\\,\\,\\forall x \\,\\,\\Rightarrow -{{\\left( x+1 \\right)}^{2}}\\le 0\\,\\,\\forall x \\\\ & \\Rightarrow B=-1-{{\\left( x+1 \\right)}^{2}}\\le -1 \\,\\,\\forall x\\\\ \\end{align}$ <br\/> V\u1eady GTLN c\u1ee7a $B$ l\u00e0 $-1 .$ <br\/> D\u1ea5u \u201c=\u201d x\u1ea3y ra khi $(x+1)^2=0\\Rightarrow x+1=0 $$\\Rightarrow x =-1 \\, ( \\text{th\u1ecfa m\u00e3n})$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 GTLN l\u00e0 $-1,$ x\u1ea3y ra khi $ x = -1$. <\/span> <br\/> <b> Ch\u00fa \u00fd:<\/b> <i>Tr\u01b0\u1edbc khi x\u00e9t GTLN c\u1ee7a m\u1ed9t bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a ph\u00e2n th\u1ee9c, ta ph\u1ea3i t\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c \u0111\u00f3. <br\/> \u0110\u1ec3 khi t\u00ecm \u0111\u01b0\u1ee3c GTLN c\u1ee7a bi\u1ec3u th\u1ee9c, x\u1ea3y ra t\u1ea1i gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a bi\u1ebfn, ta d\u1ef1a v\u00e0o \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 k\u1ebft lu\u1eadn b\u00e0i to\u00e1n.<\/i><\/span><\/span> "}],"id_ques":259},{"title":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["2"],["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai17/lv3/img\/16.jpg' \/><\/center> Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=\\dfrac{{{x}^{2}}}{x-2}\\left( \\dfrac{{{x}^{2}}+4}{x}-4 \\right)+3$ l\u00e0 _input_ <br\/> D\u1ea5u \u201c=\u201d x\u1ea3y ra khi $x = $ _input_","explain":" <span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n: $x\\ne \\{0;2\\}$ <br\/> Ta c\u00f3: <br\/> $\\begin{align} & A=\\dfrac{{{x}^{2}}}{x-2}\\left( \\dfrac{{{x}^{2}}+4}{x}-4 \\right)+3 \\\\ & =\\dfrac{{{x}^{2}}}{x-2}\\left( \\dfrac{{{x}^{2}}+4-4x}{x} \\right)+3 \\\\ & =\\dfrac{{{x}^{2}}\\left( {{x}^{2}}-4x+4 \\right)}{\\left( x-2 \\right)x}+3 \\\\ & =\\dfrac{{{x}^{2}}{{\\left( x-2 \\right)}^{2}}}{\\left( x-2 \\right)x}+3 \\\\ & =x\\left( x-2 \\right)+3 \\\\ & ={{x}^{2}}-2x+3 \\\\ & ={{x}^{2}}-2x+1+2 \\\\ & ={{\\left( x-1 \\right)}^{2}}+2 \\\\ & Do\\,\\,{{\\left( x-1 \\right)}^{2}}\\ge 0\\,\\, \\forall x\\\\ & \\Rightarrow \\,\\,A={{\\left( x-1 \\right)}^{2}}+2\\,\\,\\ge 2 \\,\\, \\forall x \\\\ \\end{align}$ <br\/> V\u1eady GTNN c\u1ee7a $A$ l\u00e0 $2$ . <br\/> D\u1ea5u \u201c=\u201d x\u1ea3y ra khi $(x-1)^2=0\\Rightarrow x-1=0 $$\\Rightarrow x =1 \\, ( \\text{th\u1ecfa m\u00e3n})$ <br\/> <span class='basic_pink'> Do \u0111\u00f3 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 GTNN l\u00e0 $2,$ x\u1ea3y ra khi $ x = 1$. <\/span> <br\/> <b> Ch\u00fa \u00fd:<\/b> <i>Tr\u01b0\u1edbc khi x\u00e9t GTNN c\u1ee7a m\u1ed9t bi\u1ec3u th\u1ee9c c\u00f3 ch\u1ee9a ph\u00e2n th\u1ee9c, ta ph\u1ea3i t\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a bi\u1ec3u th\u1ee9c \u0111\u00f3. <br\/> \u0110\u1ec3 khi t\u00ecm \u0111\u01b0\u1ee3c GTNN c\u1ee7a bi\u1ec3u th\u1ee9c, x\u1ea3y ra t\u1ea1i gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a bi\u1ebfn, ta d\u1ef1a v\u00e0o \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 k\u1ebft lu\u1eadn b\u00e0i to\u00e1n.<\/i><\/span><\/span> "}],"id_ques":260}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":44}}